Problem: Determine where $f(x)$ intersects the $x$ -axis. $f(x) = (x + 2)^2 - 36$
The function intersects the $x$ -axis where $f(x) = 0$ , so solve the equation: $ (x + 2)^2 - 36 = 0$ Add $36$ to both sides so we can start isolating $x$ on the left: $ (x + 2)^2 = 36$ Take the square root of both sides to get rid of the exponent. $ \sqrt{(x + 2)^2} = \pm \sqrt{36}$ Be sure to consider both positive and negative $6$ , since squaring either one results in $36$ $ x + 2 = \pm 6$ Subtract $2$ from both sides to isolate $x$ on the left: $ x = -2 \pm 6$ Add and subtract $6$ to find the two possible solutions: $ x = 4 \text{or} x = -8$